Question

For what value of k, the following system of equations has a unique solution : 2x + 3y – 5 = 0, kx – 6y – 8 = 0 ? 

(a) k = –4 

(b) k \(\ne\) – 4 

(c) k \(\ne\) 4 

(d) k = 4

Answer 1

(b) k \(\ne\) – 4

In the given system of equations,

a₁ = 2, b₁ = 3, c₁ = –5 

a₂ = k, b₂ = –6, c₂ = – 8

For a unique solution.

\(\frac{a_1}{a_2}\)\(\ne\)\(\frac{b_1}{b_2}\) \(\Rightarrow\) \(\frac{2}{k}\) \(\ne\) \(\frac{3}{-6}\) \(\Rightarrow\) 3k \(\ne\) -12

\(\Rightarrow\) k\(\ne\) -4